Question

Ctrl CSC 270 Final Exam-Spring2 1) [10 pts] Create anarray x which includes integers fron ltos obtain the array y which, even formula. Display both arrays x and y in a table with labeled column headings. 0 y- 2x+1 2) 120 pts] Consider the following 2D array. 01 2 -1 X-3 5 0 6 -3 7 -15 Use the standard notation and obtain the following (a) Create and display array x (b) Find the value in 2nd row and 4th column (c) Find the minimum value of first row (d) Find the sum of the elements in each row separately For each part you should have a meaningful output as below (a) X (b) value in 2nd row and 4th column: (c) minimum value in row 1: (d) sum of the elements in rowl sum of the elements in row2 sum of the elements in row3: 3) [5 pts] Create two variables called a and b. Make two pointers to a and b and call them pointer 1 and pointer 2. If a-1 and b 2 display the following (a) Value of a and memory address of a (b) Value of b and memory address of b (c) Value at pointer 1 and memory address of pointer 1 (d) Value at pointer 2 and memory address of pointer 2 4) [5 pts] Ask the user to enter few integers. User can enter any positive integer(odd or even or greater than 10). Create a stack of even integers which are between 1- 10. Display the stack.
need question 3 assap using cin and cout outputs please

Answer 1

Executable code:

#include <iostream>
using namespace std;

int main() {
   // your code goes here
   int a,b; //initializing a and b
   cin>>a>>b; //taking input of a and b
   int *pointer1, *pointer2; //initializing pointers
   //assigning values to pointers
   pointer1 = &a;
   pointer2 = &b;
   //printing values and addresses
   cout<<"value of a= "<<a<<"\nmemory address of a= "<<&a<<endl;
   cout<<"value of b= "<<b<<"\nmemory address of b= "<<&b<<endl;
   cout<<"value at pointer1="<<*pointer1<<endl;
   cout<<"nmemory address of pointer1= "<<&pointer1<<endl;
   cout<<"value at pointer2="<<*pointer2<<endl;
   cout<<"nmemory address of pointer2= "<<&pointer2<<endl;
   return 0;
}

output:

value of a= 1
memory address of a= 0x7ffc6a1f0338
value of b= 2
memory address of b= 0x7ffc6a1f033c
value at pointer1=1
nmemory address of pointer1= 0x7ffc6a1f0340
value at pointer2=2
nmemory address of pointer2= 0x7ffc6a1f0348

Running code screenshot :

12 <stdio.h> include 13 14 void DoIt( int n) 15 if( n.. 0 ) 16 printf( "DoneIn") else DoIt( n/2) printf("%d\n", n ); 20 21 23 int main() 24 DoIt( 7) return 0 25 26 27 28 29 27:0 Run Custom Input Open File Status Successfully executed Date 2019-05-16 01:40:00 Time 0 sec Mem 9.416 kB Output Done

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